Given $x+2 y>1$ $\begin{array}{l} \Rightarrow 2 y>1-x \\ \Rightarrow y>\frac{1}{2}-\frac{x}{2} \end{array}$ Consider the equation $y=\frac{1}{2}-\frac{x}{2}$ Finding points on the...
and Model 
. Model takes 6 man-hours to make per unit, while Model Y takes 10 man-hours per unit. There is a total of 450 man-hour available per week. Handling and Marketing costs are Rs 2000 and Rs 1000 per unit for Models 
and 
respectively. The total funds available for these purposes are Rs 80,000 per week. Profits per unit for Models 
and 
are Rs 1000 and Rs 500 , respectively. How many bikes of each model should the manufacturer produce so as to yield a maximum profit? Find the maximum profit.Solution: Let us take $\mathrm{x}$ an $\mathrm{y}$ to be the no. of models of bike produced by the manufacturer. From the question we have, Model $x$ takes 6 man-hours to make per unit Model $y$...
Solution: It is seen from the given figure, that the corner points are as follows: $\mathrm{R}(7 / 2,3 / 4), \mathrm{Q}(3 / 2,15 / 4), \mathrm{P}(3 / 13,24 / 13)$ and $\mathrm{S}(18 / 7,2 / 7)$ On...
subject to the constraints: 
.Solution: It is given that: $\mathrm{Z}=13 \mathrm{x}-15 \mathrm{y}$ and the constraints $\mathrm{x}+\mathrm{y} \leq 7$, $2 x-3 y+6 \geq 0, x \geq 0, y \geq 0$ Taking $x+y=7$, we have...
, subject to the constraints: 
Solution: It is given that: $Z=3 x+4 y$ and the constraints $x+y \leq 1, x \geq 0$ $\mathrm{y} \geq 0$ Taking $x+y=1$, we have $$\begin{tabular}{|l|l|l|} \hline$x$ & 1 & 0 \\ \hline$y$ & 0 & 1 \\...
subject to the constraints: 
Solution: It is given that: $\mathrm{Z}=11 \mathrm{x}+7 \mathrm{y}$ and the constraints $2 \mathrm{x}+\mathrm{y} \leq 6, \mathrm{x} \leq 2, \mathrm{x} \geq 0, \mathrm{y} \geq 0$ Let $2 x+y=6$...
and 
are at right angles.Solution: The standard form of a pair of Cartesian lines is:...
Let’s consider E to be the event of getting even number on tossing a die.
Let’s consider X to be the random variable such that X = \[0,1,2\] Now, let E = the event of drawing an ace And, F = the event of drawing non – ace So,
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Given, n coins are two headed coins and the remaining \[(n+1)\] coins are fair. Let \[{{E}_{1}}\] : the event that unfair coin is selected \[{{E}_{2}}\] : the event that the fair coin is selected...
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The probability distribution of random variable X is given by: We know that \[\sum\limits_{i=1}^{n}{P({{X}_{i}})=1}\] So, \[k\text{ }+\text{ }4k\text{ }+\text{ }9k\text{ }+\text{ }8k\text{ }+\text{...
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(i) We know that: (ii) Now, the distribution becomes \[E({{X}^{2}})\text{ }=\text{ }1\text{ }\times \text{ 1/2 }+\text{ }4\text{ }\times \text{ }1/5\text{ }+\text{ }16\text{ }\times 3/25\text{...
(i) Given, \[P\left( X\text{ }=\text{ }x \right)\text{ }=\text{ }k\left( x\text{ }+\text{ }1 \right)\]for \[x\text{ }=\text{ }1,\text{ }2,\text{ }3,\text{ }4\] So, \[P\left( X\text{ }=\text{ }1...
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Let’s consider: \[{{E}_{1}}\] = The event that the item is manufactured on machine A \[{{E}_{2}}\] = The event that the item is manufactured on machine B \[{{E}_{3}}\] = The event that the item is...
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Let \[{{E}_{1}}\] = Event that a person has TB \[{{E}_{2}}\] = Event that a person does not have TB And H = Event that the person is diagnosed to have TB. So, \[P({{E}_{1}})\text{ }=\text{...
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Given, we have \[3\] urns: Urn \[1\] = \[2\] white and \[3\] black balls Urn \[2\] = \[3\] white and 2 black balls Urn \[3\] = \[4\] white and \[1\] black balls Now, the probabilities of choosing...
Let \[{{E}_{1}}\] be the event of selecting Bag \[1\] and \[{{E}_{2}}\] be the event of selecting Bag \[2\]. Also, let \[{{E}_{3}}\] be the event that black ball is selected Now,...
Let E1 be the event that the letter comes from TATA NAGAR, E2 be the event that the letter comes from CALCUTTA And, E3 be the event that on the letter, two consecutive letters TA are visible Now,...
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Given that: \[{{A}_{1}}:\text{ }{{A}_{2}}:\text{ }{{A}_{3}}~=\text{ }4:\text{ }4:\text{ }2\] So, the probabilities will be \[P({{A}_{1}})\text{ }=\text{ }4/10,\text{ }P({{A}_{2}})\text{ }=\text{...
Referring the Question 41, here we will use Baye’s Theorem Therefore, the required probabilities are \[2/11\] and \[9/11\].
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Given: Bag \[1:3\] red balls, Bag \[2:2\] red balls and \[1\] white ball Bag \[3:3\] white balls Now, let E1, E2 and E3 be the events of choosing Bag \[1\], Bag \[2\] and Bag \[3\] respectively and...
Solution: It is given that Let's calculate the vector form: The vector eq. of as line which passes through two points whose position vectors are $\vec{a}$ and $\vec{b}$ is...
Let’s consider A to be the event of having m white and n black balls \[{{E}_{1}}\] = First ball drawn of white colour \[{{E}_{2}}\] = First ball drawn of black colour \[{{E}_{3}}\] = Second ball...
![\[\mathbf{A}\text{ }=\text{ }\left\{ \left( x,~y \right)\text{ }:~x~+~y~=\text{ }\mathbf{11} \right\}\text{ }\mathbf{B}\text{ }=\text{ }\left\{ \left( x,~y \right)\text{ }:~x~{}^\text{1}\text{ }\mathbf{5} \right\}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-885675871c9e120053f074fc2dfbccab_l3.png "Rendered by QuickLaTeX.com")
Given, two events A and B are independent such that \[\mathbf{A}\text{ }=\text{ }\left\{ \left( x,~y \right)\text{ }:~x~+~y~=\text{ }\mathbf{11} \right\}\text{ }\mathbf{B}\text{ }=\text{ }\left\{...
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Solution: Let’s take A1 to be the event of getting a total of \[6\] \[{{A}_{1}}~=\text{ }\left\{ \left( 2,\text{ }4 \right),\text{ }\left( 4,\text{ }2 \right),\text{ }\left( 1,\text{ }5...
We know that, Variance(X) = \[E({{X}^{2}})\text{ }\text{ }{{\left[ E\left( X \right) \right]}^{2}}\] Therefore, the required variance is \[665/324\].
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![\[\mathbf{P}\left( \mathbf{X}\text{ }=\text{ }\mathbf{0} \right)\text{ }=\text{ }\mathbf{P}\text{ }\left( \mathbf{X}\text{ }=\text{ }\mathbf{1} \right)\text{ }=~p~\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-75b90be1f325daf60ead8d2bd34b18b7_l3.png "Rendered by QuickLaTeX.com")
![\[\mathbf{E}({{\mathbf{X}}^{\mathbf{2}}})\text{ }=\text{ }\mathbf{E}\left[ \mathbf{X} \right]\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-f884261ac4c1cb9018e81987c5d6e5e7_l3.png "Rendered by QuickLaTeX.com")
Given, \[X\text{ }=\text{ }0,\text{ }1,\text{ }2\]and \[P\left( X\text{ }=\text{ }0 \right)\text{ }=\text{ }P\text{ }\left( X\text{ }=\text{ }1 \right)\text{ }=~p\] Let P(X) at \[X\text{ }=\text{...
Let X be the random variable scores when a die is thrown twice. \[X\text{ }=\text{ }1,\text{ }2,\text{ }3,\text{ }4,\text{ }5,\text{ }6\] And \[S\text{ }=\text{ }\left\{ \left( 1,\text{ }1...
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![\[\mathbf{P}\left[ r~\text{£}~p|s~\text{£}~p \right]\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-5dd30e9e2968f763f8ba34dc8876eb03_l3.png "Rendered by QuickLaTeX.com")
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Given, \[\mathbf{S}=\left\{ \mathbf{1},\text{ }\mathbf{2},\text{ }\mathbf{3},\text{ }\ldots .,~n \right\}\] So, \[P\left( r\text{ }\le \text{ }p/s\text{ }\le \text{ }p \right)\text{ }=\text{...
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Let’s assume \[{{E}_{1}}\] = The event that a person selected is of blood group O \[{{E}_{2}}\] = The event that the people selected is of other group And H = The event that selected person is left...
Let’s consider \[{{E}_{1}}\] = Event that the coin is fair \[{{E}_{2}}\] = Event that the coin is \[2\] headed And H = Event that the tossed coin gets head. Now, \[P({{E}_{1}})\text{ }=\text{...
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Let’s assume X to be the random variable denoting a bulb to be defective. Here, \[n\text{ }=\text{ }10,\text{ }p\text{ }=\text{ }1/50,\text{ }q\text{ }=\text{ }1\text{ }\text{ }1/50\text{ }=\text{...
![\[\mathbf{E}({{\mathbf{Y}}^{\mathbf{2}}})\text{ }=\text{ }\mathbf{2}\text{ }\mathbf{E}\left( \mathbf{X} \right)\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-4a5a28d43937af922c00cd90c0779f08_l3.png "Rendered by QuickLaTeX.com")
The probability distribution of random variable X is We know that, \[E(X)=\sum\limits_{i=1}^{n}{{{P}_{i}}{{X}_{i}}}\] \[E\left( X \right)\text{ }0.\text{ }1/5\text{ }+\text{ }1.\text{ }2/5\text{...
![\[\mathbf{P}\left( \mathbf{6} \right)\text{ }=\text{ }\mathbf{1}/\mathbf{2}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-d706b45772502fefb7902470436e85d1_l3.png "Rendered by QuickLaTeX.com")
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Therefore, the required probability distribution is
Here, we have \[X\text{ }=\text{ }0,\text{ }1,\text{ }2,\text{ }3\] [Since, die is thrown \[3\] times] And \[p\text{ }=\text{ }1/6,\text{ }q\text{ }=\text{ }5/6\] Therefore, the required expression...
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Here, random variable \[X\text{ }=\text{ }0,\text{ }1,\text{ }2\] \[P\left( X\text{ }=\text{ }2 \right)\text{ }=\text{ }P\left( 4 \right).P\left( 4 \right)\text{ }=\text{ }1/10\text{ }x\text{...
We know that: Standard deviation (S.D.) = \[\sqrt{Variance}\] So, \[Var\left( X \right)\text{ }=\text{ }E({{X}^{2}})\text{ }\text{ }{{\left[ E\left( X \right) \right]}^{2}}\] \[E\left( X...
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![\[\mathbf{P}\left( \mathbf{X}\text{ }\text{£}\text{ }\mathbf{2} \right)\text{ }+\text{ }\mathbf{P}\text{ }\left( \mathbf{X}\text{ }>\text{ }\mathbf{2} \right)\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-b2408d7c2c095809b69385b1a8434f7f_l3.png "Rendered by QuickLaTeX.com")
(i) W.k.t \[P\left( 0 \right)\text{ }+\text{ }P\left( 1 \right)\text{ }+\text{ }P\left( 2 \right)\text{ }+\text{ }P\left( 3 \right)\text{ }=\text{ }1\] \[\Rightarrow k\text{ }+\text{ }k/2\text{...
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Given We know that: \[\mathbf{Var}\left( \mathbf{X} \right)\text{ }=\text{ }\mathbf{E}({{\mathbf{X}}^{\mathbf{2}}})\text{ }\text{ }{{\left[ \mathbf{E}\left( \mathbf{X} \right)...
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Given: Total number of watches = \[100\] and number of defective watches = \[10\] So, the probability of selecting a detective watch = \[10/100\text{ }=\text{ }1/10\] Now, n = \[8\], \[p\text{...
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Here, we have n = \[7\], \[p\text{ }=\text{ }0.25\text{ }=\text{ }25/100\text{ }=\text{ 1/4}\] and \[q=1-1/4=3/4\] \[P\left( X\text{ }\ge \text{ }2 \right)\text{ }=\text{ }1\text{ }\text{ }\left[...
Here we have, \[n=10\], \[p=1/2\] and \[q=1-1/2=1/2\] \[P\left( X\text{ }\ge \text{ }8 \right)\text{ }=\text{ }P\left( x\text{ }=\text{ }8 \right)\text{ }+\text{ }P\left( x\text{ }=\text{ }9...
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Here, \[p\text{ }=\text{ }1/6\text{ }+\text{ }1/6\text{ }+\text{ }1/6\text{ }=\text{ 1/2}\Rightarrow q\text{ }=\text{ }1\text{ }\text{ 1/2 }=\text{ 1/2}\] and \[n=5\] Now, \[P\left( x\text{ }=\text{...
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Let \[{{E}_{1}},{{E}_{2}},{{E}_{3}}\] and \[{{E}_{4}}\] be the events that first, second, third and fourth card is King respectively. Therefore, the required probability is \[1/27075\].
Given that the box has \[5\] blue and \[4\] red balls. Let us consider \[{{E}_{1}}\] be the event that first ball drawn is blue and \[{{E}_{2}}\] be the event that first ball drawn is red. And, E is...
According to the question: Bag \[1\] has \[3\]B, \[2\]W balls and Bag \[2\] has \[2\]B, \[4\]W balls. Let \[{{E}_{1}}\] = The event that bag \[1\] is selected \[{{E}_{2}}\] = The event that bag...
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Let us consider \[{{W}_{1}}\] and \[{{W}_{2}}\] to be two bags containing \[\left( 4W,\text{ }5B \right)\]and \[\left( 9W,\text{ }7B \right)\]balls respectively. Let us take \[{{E}_{1}}\] be the...
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Let’s take X to be the random variable where X = \[0,500,2000\]and \[3000\]
Given that the dice is thrown three times So, the sample space n(S) = \[{{6}^{3}}~=\text{ }216\] Let E1 be the event when the sum of number on the dice was \[6\] and \[{{E}_{2}}\]be the event when...
Let’s take X to be the random variable of profit per throw. As, she loses Rs \[1\] for giving any od \[2,4,5\]. So, \[P\left( X\text{ }=\text{ }-1 \right)\text{ }=\text{ }1/6\text{ }+\text{...
Given, \[X\text{ }=\text{ }0,\text{ }1,\text{ }2,\text{ }3\] P(X = r) \[={{~}^{n}}{{C}_{r}}~{{p}^{r}}~{{q}^{n-r}}\] Where \[n\text{ }=\text{ }3,\text{ }p\text{ }=\text{ 1/2},\text{ }q\text{ }=\text{...
![\[\mathbf{P}\left( \mathbf{A}\text{ }\mathbf{B} \right)\text{ }+\text{ }\mathbf{P}\left( \mathbf{A}~B~ \right)\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-891f8144a3d214e2468e25bfa6c9744a_l3.png "Rendered by QuickLaTeX.com")
![\[\mathbf{P}\left( \mathbf{A\grave{E}}\text{ }\mathbf{B} \right)\text{ }=\text{ }\mathbf{P}\left( \mathbf{A}\text{ }\mathbf{B} \right)\text{ }+\text{ }\mathbf{P}\left( \mathbf{A}~B~ \right)\text{ }+\text{ }\mathbf{P}\left( ~A~\mathbf{}\text{ }\mathbf{B} \right)\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-503b374a6572f014a7922e0698acabb6_l3.png "Rendered by QuickLaTeX.com")
For a probability distribution, we know that if \[{{P}_{i}}~\ge \text{ }0\]
![\[p({{\mathbf{E}}_{\mathbf{1}}})\text{ }=~{{p}_{\mathbf{1}}}~\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-d36a1b249458751ac0b7265c3f2feae0_l3.png "Rendered by QuickLaTeX.com")
![\[\mathbf{P}({{\mathbf{E}}_{\mathbf{2}}})\text{ }=\text{ }{{\mathbf{p}}_{\mathbf{2}}}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-9e51eb6f05253a43a1f8b33827d4a78c_l3.png "Rendered by QuickLaTeX.com")
![\[\left( \mathbf{i} \right)~{{p}_{\mathbf{1}}}~{{p}_{\mathbf{2}}}~\left( \mathbf{ii} \right)\text{ }(\mathbf{1}{{p}_{\mathbf{1}}})~{{p}_{\mathbf{2}}}~\left( \mathbf{iii} \right)\text{ }\mathbf{1}\text{ }\text{ }(\mathbf{1}\text{ }~{{p}_{\mathbf{1}}})(\mathbf{1}\text{ }~{{p}_{\mathbf{2}}})\text{ }\left( \mathbf{iv} \right)~{{p}_{\mathbf{1}}}~+~{{p}_{\mathbf{2}}}~\text{ }\mathbf{2}{{p}_{\mathbf{1}}}{{p}_{\mathbf{2}}}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-9ab664e4b24993ddb3c0cb24b5193ff7_l3.png "Rendered by QuickLaTeX.com")
Here, \[p({{\mathbf{E}}_{\mathbf{1}}})\text{ }=~{{p}_{\mathbf{1}}}~\] and \[\mathbf{P}({{\mathbf{E}}_{\mathbf{2}}})\text{ }=\text{ }{{\mathbf{p}}_{\mathbf{2}}}\] Now, its clearly seen that either...
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![\[\mathbf{P}\left( \mathbf{A}\text{ }\mathbf{C} \right)\text{ }=\text{ }\mathbf{1}/\mathbf{5}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-b881a1a9b199441313766b0526805d42_l3.png "Rendered by QuickLaTeX.com")
![\[\mathbf{P}\left( \mathbf{B}\text{ }\mathbf{}\text{ }\mathbf{C} \right)\text{ }=\text{ 1/4}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-b3887f38cc88ac2b2b0c43663de29501_l3.png "Rendered by QuickLaTeX.com")
![\[\mathbf{P}\left( \mathbf{C}\text{ }|\text{ }\mathbf{B} \right)\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-3d783de9e6bdfea383b4c74d246a12e3_l3.png "Rendered by QuickLaTeX.com")
![\[\mathbf{P}\left( \mathbf{A}\mathbf{}\text{ }\mathbf{C} \right)\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-67318d5897bac72c39ae64c15de35bf0_l3.png "Rendered by QuickLaTeX.com")
Given, P(A) = \[2/5\], P(B) = \[1/3\] and P(C) = \[1/2\] \[\mathbf{P}\left( \mathbf{A}\text{ }\mathbf{C} \right)\text{ }=\text{ }\mathbf{1}/\mathbf{5}\]and \[\mathbf{P}\left( \mathbf{B}\text{...
![\[\mathbf{P}\left( \mathbf{A}\text{ }\mathbf{}\text{ }\mathbf{B} \right)\text{ }=\text{ }\mathbf{1}/\mathbf{4}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-642e342b173f9c41e1e70f8fabe35160_l3.png "Rendered by QuickLaTeX.com")
![\[\mathbf{P}\left( \mathbf{A}|\mathbf{B} \right)\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-437326d72ca9d40fee555f188950008d_l3.png "Rendered by QuickLaTeX.com")
![\[\mathbf{P}\left( \mathbf{B}|\mathbf{A} \right)\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-22585687a92885988b9131abed2c4ed2_l3.png "Rendered by QuickLaTeX.com")
Given, P(A) = \[1/2\], P(B) = \[1/3\] and \[\mathbf{P}\left( \mathbf{A}\text{ }\mathbf{}\text{ }\mathbf{B} \right)\text{ }=\text{ }\mathbf{1}/\mathbf{4}\] \[P\left( A \right)\text{ }=\text{ }1\text{...
As the random variable X takes \[0,1,2,3,......n\] is said to be binomial distribution having parameters n and p, if the probability is given by P(X = r) = \[^{n}{{C}_{r~}}{{p}^{r~}}{{q}^{n-r}}\]...
If two dice are thrown together, we have n(S) = \[36\] Now, let’s consider: E = A total of \[4\text{ }=\text{ }\left\{ \left( 2,\text{ }2 \right),\text{ }\left( 1,\text{ }3 \right),\text{ }\left(...
Let red marbles be presented with R and black marble with B. Also, let E be the event that at least one of the three marbles drawn be black when the first marble is red. Now, the following three...
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![\[P(\overline{A})+P(\overline{B})\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-311b0ebecaf58a7477dc7c2516ef1eca_l3.png "Rendered by QuickLaTeX.com")
W.k.t, \[A\cup B\] denotes that atleast one of the events occurs and \[A\cap B\] denotes that two events occur simultaneously. Therefore, the required answer is \[1.1\].
According to the solution of exercise 1, we have \[A\text{ }=\text{ }\left\{ \left( 1,\text{ }1 \right),\text{ }\left( 2,\text{ }2 \right),\text{ }\left( 3,\text{ }3 \right),\text{ }\left( 4,\text{...
![\[\mathbf{P}\left( \mathbf{1} \right)\text{ }=\text{ }\mathbf{P}\left( \mathbf{2} \right)\text{ }=\text{ }\mathbf{0}.\mathbf{2},\text{ }\mathbf{P}\left( \mathbf{3} \right)\text{ }=\text{ }\mathbf{P}\left( \mathbf{5} \right)\text{ }=\text{ }\mathbf{P}\left( \mathbf{6} \right)\text{ }=\text{ }\mathbf{0}.\mathbf{1}\text{ }\mathbf{and}\text{ }\mathbf{P}\left( \mathbf{4} \right)\text{ }=\text{ }\mathbf{0}.\mathbf{3}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-4174be6da398129f0164c8571610db50_l3.png "Rendered by QuickLaTeX.com")
Given that a loaded die is thrown such that \[\mathbf{P}\left( \mathbf{1} \right)\text{ }=\text{ }\mathbf{P}\left( \mathbf{2} \right)\text{ }=\text{ }\mathbf{0}.\mathbf{2},\text{ }\mathbf{P}\left(...
Solution: The points $(4,7,8),(2,3,4)$ and $(-1,-2,1),(1,2,5)$. Consider $A B$ be the line joining the points, $(4,7,8),(2,3,4)$ and $C D$ be the line through the points $(-1,-2$, 1), $(1,2,5)$. So...
As per the given question, Let I = = = = = = = = =
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Let’s take x an y to be the number of models of bike produced by the manufacturer. From the question we have, Model x takes \[6\] man-hours to make per unit Model y takes \[10\] man-hours to make...
![\[x~+\text{ }\mathbf{4}y~\text{£}\text{ }\mathbf{8},\text{ }\mathbf{2}x~+\text{ }\mathbf{3}y~\text{£}\text{ }\mathbf{12},\text{ }\mathbf{3}x~+~y~\text{£}\text{ }\mathbf{9},~x~{}^\text{3}\text{ }\mathbf{0},~y~{}^\text{3}\text{ }\mathbf{0}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-979bb74f9bdb51824148746fc9c40530_l3.png "Rendered by QuickLaTeX.com")
Given: Z = x + y subject to constraints, \[x~+\text{ }\mathbf{4}y~\text{£}\text{ }\mathbf{8},\text{ }\mathbf{2}x~+\text{ }\mathbf{3}y~\text{£}\text{ }\mathbf{12},\text{...
As per the solution of exercise 15, we have Maximize Z = x + y, subject to the constraints \[2x\text{ }+\text{ }3y\le 120\] … (i) \[8x\text{ }+\text{ }5y\le 400\]… (ii) \[x\ge 0,\text{ }y\ge 0\]...
As per the solution of exercise 14, we have Maximize \[Z\text{ }=\text{ }200x\text{ }+\text{ }120y\]subject to constrains \[3x\text{ }+\text{ }y\le 600\]…. (i) \[x\text{ }+\text{ }y\le 300\]…. (ii)...
From the solution of exercise 13, we have The objective function for maximum profit \[Z\text{ }=\text{ }100x\text{ }+\text{ }170y\] Subject to constraints, \[x\text{ }+\text{ }4y\le 1800\]…. (i)...
As per the solution of exercise 12, we have The objective function for minimum cost is \[Z\text{ }=\text{ }400x\text{ }+\text{ }200y\] Subject to the constrains; \[5x\text{ }+\text{ }2y\ge 30\]….....
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As per the solution of exercise \[11\], we have Maximize \[Z\text{ }=\text{ }50x\text{ }+\text{ }60y\]subject to the constraints \[20x\text{ }+\text{ }10y\le 200\text{ }2x\text{ }+\text{ }y\le 20\]…...
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Let’s assume the man covers x km on his motorcycle at the speed of \[50\]km/hr and covers y km at the speed of \[50\] km/hr and covers y km at the speed of \[80\] km/hr. So, cost of petrol =...
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Let’s assume x and y to be the number of sweaters of type A and type B respectively. From the question, the following constraints are: \[360x\text{ }+\text{ }120y\le 72000\Rightarrow 3x\text{...
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Let’s consider that the company manufactures x boxes of type A screws and y boxes of type B screws. From the given information the below table is constructed: From the data in the above table, the...
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Let us consider x and y to be the number of large and small vans respectively. From the given information the below constrains table is constructed: Now, the objective function for minimum cost is...
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Let x units of type A and y units of type B electric circuits be produced by the manufacturer. From the given information the below table is constructed: Now, the total profit function in rupees...
![\[\mathbf{Z}\text{ }=~x~+\text{ }\mathbf{2}y\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-b80ed52b8ae39a40c6089b682f75c471_l3.png "Rendered by QuickLaTeX.com")
From the given figure, it’s seen that the corner points are as follows: \[R\left( 7/2,\text{ }3/4 \right),\text{ }Q\left( 3/2,\text{ }15/4 \right),\text{ }P\left( 3/13,\text{ }24/13 \right)\text{...
![\[\mathbf{Z}\text{ }=\text{ }\mathbf{4}x~+~y\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-552a20e496431cb5a75822a3ece9d00c_l3.png "Rendered by QuickLaTeX.com")
According to the question: \[\mathbf{Z}\text{ }=\text{ }\mathbf{4}x~+~y\] In the given figure, ABC is the feasible region which is open unbounded. Here, we have \[x\text{ }+\text{ }y\text{ }=\text{...
In the evaluating table for the value of Z, it’s clearly seen that the maximum value of Z is \[47\] at \[(3,2)\]
![\[\mathbf{Z}\text{ }=\text{ }\mathbf{11}x~+\text{ }\mathbf{7}y\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-5c7a02563203520b7505fe3ddb709e9f_l3.png "Rendered by QuickLaTeX.com")
In the given figure, it’s seen that the feasible region is ABCA. The corner points are \[C\left( 0,\text{ }3 \right),\text{ }B\left( 0,\text{ }5 \right)\]’and for A, we have to solve equations...
![\[\mathbf{Z}\text{ }=\text{ }\mathbf{5}x~+\text{ }\mathbf{7}y\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-55e25535c515d2a5fba086ba449805a2_l3.png "Rendered by QuickLaTeX.com")
According to the question: \[\mathbf{Z}\text{ }=\text{ }\mathbf{5}x~+\text{ }\mathbf{7}y\] and feasible region OABC. The corner points of the feasible region are \[O\left( 0,\text{ }0 \right),\text{...
![\[\mathbf{Z}\text{ }=\text{ }\mathbf{3}x~+\text{ }\mathbf{4}y~\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-390dc45e23eba82376da394c50e1be8b_l3.png "Rendered by QuickLaTeX.com")
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As shown in the figure, OAED is the feasible region. At A, \[y\text{ }=\text{ }0\]in equation \[2x\text{ }+\text{ }y\text{ }=\text{ }104\]we get, \[x=52\] This is a corner point \[A\text{ }=\text{...
![\[\mathbf{Z}\text{ }=\text{ }\mathbf{13}x~\text{ }\mathbf{15}y~\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-338f18452f410c945d5cabf1924ea645_l3.png "Rendered by QuickLaTeX.com")
![\[x~+~y~\text{£}\text{ }\mathbf{7},\text{ }\mathbf{2}x~\text{ }\mathbf{3}y~+\text{ }\mathbf{6}\text{ }{}^\text{3}\text{ }\mathbf{0},~x~{}^\text{3}\text{ }\mathbf{0},~y~{}^\text{3}\text{ }\mathbf{0}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-e55d23f4fcb872e75b4a8e7d4f615d75_l3.png "Rendered by QuickLaTeX.com")
According to the question: \[\mathbf{Z}\text{ }=\text{ }\mathbf{13}x~\text{ }\mathbf{15}y~\]and the constraints \[x~+~y~\text{£}\text{ }\mathbf{7},\text{ }\mathbf{2}x~\text{ }\mathbf{3}y~+\text{...
![\[x~\text{£}\text{ }\mathbf{3},~y~\text{£}\text{ }\mathbf{2},~x~{}^\text{3}\text{ }\mathbf{0},~y~{}^\text{3}\text{ }\mathbf{0}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-32897c7ada69b747fc2df4430eae78b1_l3.png "Rendered by QuickLaTeX.com")
According to the question: \[\mathbf{Z}\text{ }=\text{ }\mathbf{11}x~+\text{ }\mathbf{7}y\]and the constraints \[x~\text{£}\text{ }\mathbf{3},~y~\text{£}\text{ }\mathbf{2},~x~{}^\text{3}\text{...
![\[\mathbf{Z}\text{ }=\text{ }\mathbf{3}x~+\text{ }\mathbf{4}y\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-bb613b36107ec1e7348202194fd16fe6_l3.png "Rendered by QuickLaTeX.com")
![\[x~+~y~\text{£}\text{ }\mathbf{1},~x~{}^\text{3}\text{ }\mathbf{0},~y~{}^\text{3}\text{ }\mathbf{0}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-8445e8a11a750a4c9e5f61100f78a870_l3.png "Rendered by QuickLaTeX.com")
According to the question: \[\mathbf{Z}\text{ }=\text{ }\mathbf{3}x~+\text{ }\mathbf{4}y\] and the constraints \[x~+~y~\text{£}\text{ }\mathbf{1},~x~{}^\text{3}\text{...
![\[\mathbf{Z}\text{ }=\text{ }\mathbf{11}x~+\text{ }\mathbf{7}y~\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-a04a1e74298ed4c91cc33f681051634b_l3.png "Rendered by QuickLaTeX.com")
![\[\mathbf{2}x~+~y~\text{£}\text{ }\mathbf{6},~x~\text{£}\text{ }\mathbf{2},~x~{}^\text{3}\text{ }\mathbf{0},~y~{}^\text{3}\text{ }\mathbf{0}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-3bfc9eae4e5c2facd047b27f9d3525c4_l3.png "Rendered by QuickLaTeX.com")
According to the question: \[\mathbf{Z}\text{ }=\text{ }\mathbf{11}x~+\text{ }\mathbf{7}y~\]and the constraints \[\mathbf{2}x~+~y~\text{£}\text{ }\mathbf{6},~x~\text{£}\text{...
dx/4+9x^2 (A) pi/6 (B) pi/12 (C) pi/24 (D) pi/4= = = = = = Therefore, The correct option is option (C) .
According to ques,
(
)z=0According to ques, planes are: \[~ax\text{ }+\text{ }by\text{ }=\text{ }0\text{ }\ldots .\text{ }\left( i \right)\] and \[z\text{ }=\text{ }0\text{ }\ldots .\text{ }\left( ii \right)\] hence, the...
According to ques, planes are \[{{P}_{1}}:\text{ }5x\text{ }+\text{ }3y\text{ }+\text{ }6z\text{ }+\text{ }8\text{ }=\text{ }0\] \[{{P}_{2}}:\text{ }x\text{ }+\text{ }2y\text{ }+\text{ }3z\text{...
According to the ques,
According to ques, Equation of the plane passing through two points, \[\left( {{x}_{1}},\text{ }{{y}_{1}},\text{ }{{z}_{1}} \right)\text{ }and\text{ }\left( {{x}_{2}},\text{ }{{y}_{2}},\text{...
According to ques, Point is \[\left( 3,\text{ }0,\text{ }1 \right)\] and the equation of planes are: \[x\text{ }+\text{ }2y\text{ }=\text{ }0\text{ }\ldots .\text{ }\left( i \right)\] and \[3y\text{...
tanx dx= = = = = =
According to ques, plane is: \[2x\text{ }\text{ }2y\text{ }+\text{ }4z\text{ }+\text{ }5\text{ }=\text{ }0\] and point: \[\left( 1,\text{ }3/2,\text{ }2 \right)\] The direction ratios of the...
According to ques,
according to ques, The coordinates of any point Q on the line are: \[x\text{ }=\text{ }-2\lambda \text{ }+\text{ }4,\text{ }y\text{ }=\text{ }6\lambda \text{ }and\text{ }z\text{ }=\text{ }-3\lambda...
Let, \[~OX,\text{ }OY,\text{ }OZ\text{ }and\text{ }ox,\text{ }oy,\text{ }oz\] to be two rectangular systems According to ques,
According to ques, \[~O\text{ }\left( 0,\text{ }0,\text{ }0 \right)\text{ }and\text{ }A\left( a,\text{ }b,\text{ }c \right)\] Hence, the direction ratios of OA : \[a\text{ }\text{ }0,\text{ }b\text{...
According to ques, \[~l,\text{ }m,\text{ }n\text{ }and~l~+\text{ }dl,~m~+\text{ }dm,~n~+\text{ }dn\] are the direction cosines of a variable line in two positions hence, \[{{l}^{2}}~+\text{...
According to ques, Equations are, \[l~+~m~+~n~=\text{ }0\text{ }\ldots ..\text{ }\left( i \right)\] \[{{l}^{2}}~+~{{m}^{2}}~~{{n}^{2}}~=\text{ }0\text{ }\ldots .\text{ }\left( ii \right)\] From...
According to ques, Any point in the given line is: \[x\text{ }\text{ }3/\text{ }2\text{ }=\text{ }y\text{ }\text{ }3/1\text{ }=\text{ }z/1\text{ }=\text{ }\lambda \] Or, \[x\text{ }=\text{ }2\lambda...
According to ques, points are: \[\left( 2,\text{ }1,\text{ }0 \right),\text{ }\left( 3,\text{ }2,\text{ }2 \right)\text{ }and\text{ }\left( 3,\text{ }1,\text{ }7 \right)\] Since, equation of the...
According to ques, Points are : \[\left( 2,\text{ }\text{ }1,\text{ }\text{ }3 \right)\text{ }and\text{ }\left( 1,\text{ }\text{ }3,\text{ }3 \right)\] And, Direction ratios of the normal to the...
Since, the normal to the plane is equally inclined to the axes. so, \[cos\text{ }\alpha \text{ }=\text{ }cos\text{ }\beta \text{ }=\text{ }cos\text{ }\gamma \] or, \[So,\text{ }co{{s}^{2}}~\alpha...
According to ques, Coordinates are \[A\text{ }\left( 2,\text{ }3,\text{ }4 \right)\text{ }and\text{ }B\text{ }\left( 4,\text{ }5,\text{ }8 \right)\] Again, Coordinates of the mid-point C are...
According to ques,
According to ques, Points are \[~A\text{ }\left( 0,\text{ }1,\text{ }1 \right)\text{ }and\text{ }B\text{ }\left( 4,\text{ }5,\text{ }1 \right)\] \[C\text{ }\left( 3,\text{ }9,\text{ }4 \right)\text{...
(2i+j+2k) and r=(2j-5k)+
(6i+3j+2k)According to ques,
According to ques,
According to equation of line we have,
Since, \[co{{s}^{2}}~\alpha \text{ }+\text{ }co{{s}^{2}}~\beta \text{ }+\text{ }co{{s}^{2}}~\gamma \text{ }=\text{ }1\] or, \[co{{s}^{2}}~{{60}^{o}}~+\text{ }co{{s}^{2}}~{{45}^{o}}~+\text{...
Solution: CORRECT OPTION: (B) as the ques suggests,
CORRECT OPTION: \[\left( C \right)-5i+2j+4k\] Let us take, A and B be the pints with coordinates (2, 5, 0) and (-3, 7, 4) respectively. Hence,
CORRECT ANSWER: \[\left( D \right)5a/4~\] ACCORDING TO QUES , Given ratio is 3:1
CORRECT ANSWER: \[\left( C \right)\text{ }3\left( i-2j+2k \right)\]
according to ques,
according to ques,
according to ques,
according to ques,
According to ques, proved.
according to ques,
According to ques,
according to ques,
according to ques,
according to ques,
according to ques,
According to ques,
Since, vector makes equal angles with the axes, their direction cosines will be same hence, \[~l\text{ }=\text{ }m\text{ }=\text{ }n\] also, \[{{l}^{2}}~+\text{ }{{m}^{2}}~+\text{...
according to ques,
According to ques,
According to ques,
According to ques,
according to ques,
Correct option :(B) According to ques, \[y~=\text{ }A\text{ }cos\text{ }ax~+\text{ }B\text{ }sin\text{ }ax\] Differentiating with respect to x,
Correct option :(C). According to ques, \[y~=~{{e}^{x}}~\left( A\text{ }cos~x~+\text{ }B\text{ }sin~x \right)\] Differentiating both sides with respect to x,
Correct option : (A) 2 and not defined According to ques, Since the degree of dy/dx is in fraction its undefined and the degree is \[2.\]
Correct option: D (2) According to ques, hence the answer is 2.
Correct option: (D) not defined. As the value of sin (dy/dx) on expansion will be in increasing power of dy/dx,
According to ques,
Let \[P\left( x\text{ },\text{ }y \right)\] be any point on the curve, AB be the tangent to the given curve at P. A/Q, P is the mid-point of AB HENCE, the coordinates of \[A\text{ }is\text{ }\left(...
According to the ques, The slope of the tangent of the curve \[=\text{ }dy/dx\] Now, the difference between the abscissa and ordinate \[=\text{ }x\text{ }\text{ }y\] Hence A/Q, \[dy/dx\text{...
According to ques, Since, the line is passing through the point (1, 0), then \[\left( 0\text{ }\text{ }1 \right)\text{ }\left( 1\text{ }+\text{ }1 \right)\text{ }=\text{ }c\left( 1...
ACCORDING TO QUES, Line is passing through the point \[\left( 1,\text{ }0 \right),SO,\] \[\left( 0\text{ }\text{ }1 \right)\text{ }\left( 1\text{ }+\text{ }1 \right)\text{ }=\text{ }c\left( 1...
ACCORDING TO QUES, It’s a homogeneous differential function,
According to ques, \[~P\text{ }=\text{ }-3\text{ }and\text{ }Q\text{ }=\text{ }sin\text{ }2x\]
According to ques,
according to ques,
according to the ques,
according to ques, concentric circles with centre \[\left( 1,\text{ }2 \right)\] and with radius ‘r’ can be written as, \[{{\left( x\text{ }\text{ }1 \right)}^{2}}~+\text{ }{{\left( y\text{ }\text{...
ACCORDING TO QUES, \[~\left( 1\text{ }+~{{y}^{2}} \right)\text{ }ta{{n}^{1}}x\text{ }dx~+\text{ }2y~\left( 1\text{ }+~{{x}^{2}} \right)~dy~=\text{ }0\] \[2y\text{ }\left( 1\text{ }+\text{ }{{x}^{2}}...
ACCORDING TO QUES, DIFFERENTIATING WITH RESPECT TO x ,
according to ques,
according to ques,
According to ques, or, \[\left( x~+~y \right)\text{ }\left( dx~~dy \right)\text{ }=~dx~+~dy\] \[\left( x\text{ }+\text{ }y \right)\text{ }dx\text{ }\text{ }\left( x\text{ }\text{ }y \right)\text{...
according to ques,
according to ques,
according to ques,
according to ques,
According to ques, Equation will be, \[{{\left( x\text{ }\text{ }0 \right)}^{2}}~+\text{ }{{\left( y\text{ }\text{ }a \right)}^{2}}~=\text{ }{{a}^{2}},\] where (0, a) is the centre...
according to ques,
according to ques,
according to ques,
according to ques,
according to ques, we can write,
According to ques we have, \[~ydx\text{ }\text{ }xdy\text{ }=\text{ }{{x}^{2}}ydx\] or, \[y\text{ }dx\text{ }\text{ }{{x}^{2}}y\text{ }dx\text{ }=\text{ }xdy\] \[y\text{ }\left( 1\text{ }\text{...
ACCORDING TO QUES, \[dy/dx\text{ }+\text{ }1\text{ }=\text{ }{{e}^{x+y}}\] NOW PUTTING, \[x\text{ }+\text{ }y\text{ }=\text{ }t\] and differentiating w.r.t. x,
ACCORDING TO QUES, \[dy/dx\text{ }+\text{ }ay\text{ }=\text{ }{{e}^{mx}}\] for linear differential equation of first order, \[P\text{ }=\text{ }a\text{ }and\text{ }Q\text{ }=\text{...
according to ques, \[dy/dx\text{ }+\text{ }2xy\text{ }=\text{ }y\]
according to ques,
according to ques,
SINCE ,equation of all non-vertical lines is $$ \[~y\text{ }=\text{ }mx\text{ }+\text{ }c\] On differentiating w.r.t. x, \[dy/dx\text{ }=\text{ }m\] and, on differentiating w.r.t. x,...
ACCORDING TO QUES,
Ans:
Let \[~P\left( n \right)\text{ }:\text{ }{{\left( AB \right)}^{n}}~=\text{ }{{A}^{n}}{{B}^{n}}\] In this way, \[P\left( 1 \right)\text{ }:\text{ }{{\left( AB \right)}^{1}}~=\text{...
Ans:
We realize that, \[A\text{ }.\text{ }I\text{ }=\text{ }I\text{ }.\text{ }A\] In this way, \[A\text{ }and\text{ }I\]are commutative. Subsequently, we can grow \[{{\left( I\text{ }+\text{ }A...
Ans:
Ans:
Ans: By utilizing equity of matrixs, we get \[co{{s}^{2}}~\alpha \text{ }+\text{ }si{{n}^{2~}}\alpha \text{ }=\text{ }1\], which is valid for all genuine upsides of \[\alpha .\]
Ans:
Ans: Let, Presently, by equity of matrices, we get \[a\text{ }=\text{ }1,\text{ }b\text{ }=\text{ }-\text{ }2,\text{ }c\text{ }=\text{ }-\text{ }5\] Furthermore, \[2a\text{ }\text{ }d\text{ }=\text{...
Ans: Presently, \[3a\text{ }=\text{ }a\text{ }+\text{ }4\Rightarrow a\text{ }=\text{ }2\] \[3b\text{ }=\text{ }6\text{ }+\text{ }a\text{ }+\text{ }b\] \[3b\text{ }-\text{ }b\text{ }=\text{ }8\]...
Ans:
Ans: How about we think about a matrix C, with the end goal that Looking at the terms, \[\mathbf{2a}\text{ }+\text{ }\mathbf{48}\text{ }=\text{ }\mathbf{0}\text{ }\Rightarrow \text{...
Answer: In the given matrix condition, Looking at the comparing components, we get \[x\text{ }+\text{ }y\text{ }=\text{ }6,\] \[xy\text{ }=\text{ }8,\] \[z\text{ }+\text{ }6\text{ }=\text{ }0\]and...
Ans: Since, we obtain all the zeroes in a row of the matrix A, L.H.S., A-1 does not exist.
Ans:
Ans:
Answer: According to the given ques
Ans:
Answer: According to the given ques,
Answer: According to the given ques, (a) (b)
Answer: According to the given ques
Answer: (a) (b)
Answer: According to the given ques
Given, \[A\text{ }and\text{ }B\]are square matrices with the end goal that \[AB\text{ }=\text{ }BA.\] In this way, \[{{\left( A\text{ }+\text{ }B \right)}^{2}}~=\text{ }\left( A\text{ }+\text{ }B...
As, \[A\text{ }and\text{ }B\]be square matrixs of order \[3\text{ }x\text{ }3.\] We have, \[{{\left( AB \right)}^{2}}~=\text{ }AB\text{ }.\text{ }AB\] \[=\text{ }A\left( BA \right)B\] \[=\text{...
Answer: According to the given ques
Answer: According to the given ques